Condition on a prime ideal that implies it is maximal. If $P$ is a prime ideal in a ring $R$, and if $P$ is contained in only finitely many right ideals of $R$, show that $P$ is a maximal ideal of $R$.
I'm honestly not sure where to start with this. Are there any hints?
 A: After factoring by $P$ the question reduces to the
same question with $P=0$, which can be stated as:
Show that if $R$ is a prime ring with finitely many right ideals,
then $R$ is simple. [$R$ is prime means its zero ideal is a prime
ideal, and $R$ is simple means its zero ideal is maximal.]
Reason the statement is true:
As Curious noted, having finitely many right ideals
implies that $R$ is right Artinian. This forces the Jacobson radical
$J=J(R)$ to be nilpotent.
But $R$ is prime, so any nilpotent ideal of $R$ is zero.
This implies that $R$ is Artinian and Jacobson semisimple, which
implies that $R$ is semisimple. But a prime semisimple
ring is simple by the Wedderburn-Artin Theorem.
We can go a little bit further than required.
The Wedderburn-Artin Theorem shows that $R\cong M_n(D)$
for some $n$ and some division ring $D$. But such
rings usually have infinitely many right ideals.
The exceptions are when $R$ itself is a division ring,
or $R$ is finite. So one can reword the original problem as:
Show that if $P$ is a prime ideal of $R$ that is contained
in finitely many right ideals, then either
$P$ is maximal as a right ideal in $R$ or else $P$ is
maximal as a 2-sided ideal of $R$ and $P$ has finite index in $R$.
A: If $R$ is a commutative ring, then $R/P$ is an integral domain. Since ideals in $R/P$ correspond to ideals in $R$ containing $P$, we have that $R/P$ has only finitely many ideals. Such a ring is Artinian. One can use this fact to show any nonzero element has an inverse (for example, here). Thus, $R/P$ is in fact a field, which happens if and only if $P$ is maximal, and we are done.
